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Nesterboy [21]
3 years ago
8

In comparison with liquids and gases, solids are

Chemistry
1 answer:
Citrus2011 [14]3 years ago
5 0
In comparison with liquids and gases, solids are more dense. The answer is letter B. <span>The solid has a more definite shape and volume. The particles are locked into place. It cannot be further compressed due to the bond that exists between the molecules. The kinetic energy of the molecules is close to none because the molecules are so close and so compact with each other. </span>
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22.4 L of hydrogen is taken in a cylinder at zero degree Celsius and at one atm. This gas is transferred to a cylinder of 11.2 L
geniusboy [140]

Answer:

  2 Atm; 2.016 g

Explanation:

Changing the volume without changing the temperature or mass only changes the pressure. Volume and pressure are inversely proportional so halving the volume will double the pressure.

P = 1 Atm, T = 0 °C are "standard" temperature and pressure (STP). The volume of 1 mole of gas is 22.4 L under these conditions. That means the amount of hydrogen gas in the cylinder is 1 mole, so has a mass of 2.016 g.

After the volume reduction, the pressure is 2 Atm, and the mass remains 2.016 g.

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3 years ago
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Use the following graph of a car traveling on a straight northerly path to answer this question. At what time would the
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Answer:

B 144.0 s is the best answer of this question

6 0
4 years ago
Find the hydroxide ion concentration of a HBr solution with a pH of 4.75
Sunny_sXe [5.5K]
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3 years ago
Ions can have a positive or negative charge. false or true
Fudgin [204]
True. Ions can have a positive or negative charge.
7 0
3 years ago
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A mixture of methane and air is capable of being ignited only if the mole percent of methane is between 5% and 15%. A mixture co
DaniilM [7]

Answer:

Explanation:

The air 9% mole% methane have an average molecular weight of:

9%×16,04g/mol + 91%×29g/mol = 27,8g/mol

And a flow of 700000g/h÷27,8g/mol = 25180 mol/h

In the reactor where methane solution and air are mixed:

In = Out

Air balance:

91% air×25180 mol/h + 100% air×X = 95%air×(X+25180)

Where X is the flow rate of air in mol/h = <em>20144 mol air/h</em>

<em></em>

The air in the product gas is

95%×(20144 + 25180) mol/h = 43058 mol air× 21%O₂ = 9042 mol O₂ ×32g/mol = <em>289 kg O₂</em>

43058 mol air×29g/mol <em>1249 kg air</em>

Percent of oxygen is: \frac{289kg}{1249 kg} =<em>0,231 kg O₂/ kg air</em>

<em></em>

I hope it helps!

4 0
3 years ago
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